Complex Multiplication - Elliptic Curves: Number Theory and Cryptography - page 328

and the first term for j ◦ S is

ζ −b e − 2 πiaτ/d = ζ −b ( e − 2 πiτ/N ) a 2 .

= a 2 . Therefore, these terms represent
different powers of e 2 πiτ/N , so they cannot cancel each other. One of them
must be the first term of the expansion of j − j ◦ S , which therefore has
coecient 1 or −ζ b . In particular, for each factor j − j ◦ S , the coecient of
the first term of the expansion is a root of unity. The coecient of the first
term of the expansion of H N ( j ) is the product of these roots of unity, hence
a root of unity. Also, since the terms don't cancel each other, the first term
of each factor contains a negative power of e 2 πiτ/N . Therefore, the first term
of the expansion H N ( j ) is a negative power of q ,so H N ( X )isnonconstant.
Suppose H N ( X )= uX + lower terms. We know that u

Since N is not a perfect square, N

∈ Z . Since the

Laurent series for j starts with 1 /q ,

H N ( j )= uq − +higher terms .

We have shown that u is a root of unity. Since it is an integer, u = ± 1. This
completes the proof of (2).

The modular polynomial Φ N ( X, Y ) has rather large coe cients. For

example,

X 2 Y 2 + X 3 + Y 3 +2 4

Φ 2 ( X, Y )= − · 3 · 31 XY ( X + Y )

[phone] ( X 2 + Y 2 ) · · 4027 XY − · ·

+2 8 · 3 7 · 5 6 ( X + Y ) − 2 12 · 3 9 · 5 9 ,

and

− 1069956 X 3 Y + 36864000 X 3
+2232 X 2 Y 3 + 2587918086 X 2 Y 2 + 8900222976000 X 2 Y
[phone] X 2

= X 4 − X 3 Y 3 + 2232 X 3 Y 2

Φ 3 ( X, Y )

1069956 XY 3 + 8900222976000 XY 2 −

770845966336000000 XY + 1855425871872000000000 X + Y 4
[phone] Y 3 + 452984832000000 Y 2 + 1855425871872000000000 Y

−

For Φ N for higher N , see [50], [53], [54].
We can now prove Theorem 10.9. Let R be an order in an imaginary
quadratic field and let L be a lattice with RL ⊆ L . By multiplying L by a

suitable factor, we may assume that

L = Z + Z τ

with τ ∈H . T he o rder R is o f fin ite index in O K for some imaginary quadratic

fiel d K = Q ( √ −d ). Since √ −d ∈O K , there is a nonzero integer n such that

n √ − R . Therefore, n √ − d ∈ L ,so

n √ −d · τ = tτ + u,

dL ⊆

n √ −d · 1= vτ + w (10.3)